3.37 \(\int \frac{1}{(1+\cosh ^2(x))^3} \, dx\)

Optimal. Leaf size=51 \[ \frac{19 \tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{2}}\right )}{32 \sqrt{2}}-\frac{9 \sinh (x) \cosh (x)}{32 \left (\cosh ^2(x)+1\right )}-\frac{\sinh (x) \cosh (x)}{8 \left (\cosh ^2(x)+1\right )^2} \]

[Out]

(19*ArcTanh[Tanh[x]/Sqrt[2]])/(32*Sqrt[2]) - (Cosh[x]*Sinh[x])/(8*(1 + Cosh[x]^2)^2) - (9*Cosh[x]*Sinh[x])/(32
*(1 + Cosh[x]^2))

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Rubi [A]  time = 0.0512598, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {3184, 3173, 12, 3181, 206} \[ \frac{19 \tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{2}}\right )}{32 \sqrt{2}}-\frac{9 \sinh (x) \cosh (x)}{32 \left (\cosh ^2(x)+1\right )}-\frac{\sinh (x) \cosh (x)}{8 \left (\cosh ^2(x)+1\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(1 + Cosh[x]^2)^(-3),x]

[Out]

(19*ArcTanh[Tanh[x]/Sqrt[2]])/(32*Sqrt[2]) - (Cosh[x]*Sinh[x])/(8*(1 + Cosh[x]^2)^2) - (9*Cosh[x]*Sinh[x])/(32
*(1 + Cosh[x]^2))

Rule 3184

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> -Simp[(b*Cos[e + f*x]*Sin[e + f*x]*(a + b*Sin[
e + f*x]^2)^(p + 1))/(2*a*f*(p + 1)*(a + b)), x] + Dist[1/(2*a*(p + 1)*(a + b)), Int[(a + b*Sin[e + f*x]^2)^(p
+ 1)*Simp[2*a*(p + 1) + b*(2*p + 3) - 2*b*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f}, x] && NeQ
[a + b, 0] && LtQ[p, -1]

Rule 3173

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Sim
p[((A*b - a*B)*Cos[e + f*x]*Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^(p + 1))/(2*a*f*(a + b)*(p + 1)), x] - Dist[1/
(2*a*(a + b)*(p + 1)), Int[(a + b*Sin[e + f*x]^2)^(p + 1)*Simp[a*B - A*(2*a*(p + 1) + b*(2*p + 3)) + 2*(A*b -
a*B)*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && LtQ[p, -1] && NeQ[a + b, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3181

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[1/(a + (a + b)*ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{\left (1+\cosh ^2(x)\right )^3} \, dx &=-\frac{\cosh (x) \sinh (x)}{8 \left (1+\cosh ^2(x)\right )^2}-\frac{1}{8} \int \frac{-7+2 \cosh ^2(x)}{\left (1+\cosh ^2(x)\right )^2} \, dx\\ &=-\frac{\cosh (x) \sinh (x)}{8 \left (1+\cosh ^2(x)\right )^2}-\frac{9 \cosh (x) \sinh (x)}{32 \left (1+\cosh ^2(x)\right )}-\frac{1}{32} \int -\frac{19}{1+\cosh ^2(x)} \, dx\\ &=-\frac{\cosh (x) \sinh (x)}{8 \left (1+\cosh ^2(x)\right )^2}-\frac{9 \cosh (x) \sinh (x)}{32 \left (1+\cosh ^2(x)\right )}+\frac{19}{32} \int \frac{1}{1+\cosh ^2(x)} \, dx\\ &=-\frac{\cosh (x) \sinh (x)}{8 \left (1+\cosh ^2(x)\right )^2}-\frac{9 \cosh (x) \sinh (x)}{32 \left (1+\cosh ^2(x)\right )}+\frac{19}{32} \operatorname{Subst}\left (\int \frac{1}{1-2 x^2} \, dx,x,\coth (x)\right )\\ &=\frac{19 \tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{2}}\right )}{32 \sqrt{2}}-\frac{\cosh (x) \sinh (x)}{8 \left (1+\cosh ^2(x)\right )^2}-\frac{9 \cosh (x) \sinh (x)}{32 \left (1+\cosh ^2(x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.170723, size = 51, normalized size = 1. \[ \frac{19 \tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{2}}\right )}{32 \sqrt{2}}-\frac{9 \sinh (2 x)}{32 (\cosh (2 x)+3)}-\frac{\sinh (2 x)}{4 (\cosh (2 x)+3)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Cosh[x]^2)^(-3),x]

[Out]

(19*ArcTanh[Tanh[x]/Sqrt[2]])/(32*Sqrt[2]) - Sinh[2*x]/(4*(3 + Cosh[2*x])^2) - (9*Sinh[2*x])/(32*(3 + Cosh[2*x
]))

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Maple [B]  time = 0.015, size = 129, normalized size = 2.5 \begin{align*} -{\frac{1}{4} \left ({\frac{11}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{7}}+{\frac{7}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{5}}+{\frac{7}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{3}}+{\frac{11}{8}\tanh \left ({\frac{x}{2}} \right ) } \right ) \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{4}+1 \right ) ^{-2}}+{\frac{19\,\sqrt{2}}{256}\ln \left ({ \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+\sqrt{2}\tanh \left ({\frac{x}{2}} \right ) +1 \right ) \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}-\sqrt{2}\tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}} \right ) }-{\frac{19\,\sqrt{2}}{256}\ln \left ({ \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}-\sqrt{2}\tanh \left ({\frac{x}{2}} \right ) +1 \right ) \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+\sqrt{2}\tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1+cosh(x)^2)^3,x)

[Out]

-1/4*(11/8*tanh(1/2*x)^7+7/8*tanh(1/2*x)^5+7/8*tanh(1/2*x)^3+11/8*tanh(1/2*x))/(tanh(1/2*x)^4+1)^2+19/256*2^(1
/2)*ln((tanh(1/2*x)^2+2^(1/2)*tanh(1/2*x)+1)/(tanh(1/2*x)^2-2^(1/2)*tanh(1/2*x)+1))-19/256*2^(1/2)*ln((tanh(1/
2*x)^2-2^(1/2)*tanh(1/2*x)+1)/(tanh(1/2*x)^2+2^(1/2)*tanh(1/2*x)+1))

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Maxima [A]  time = 1.71825, size = 112, normalized size = 2.2 \begin{align*} -\frac{19}{128} \, \sqrt{2} \log \left (-\frac{2 \, \sqrt{2} - e^{\left (-2 \, x\right )} - 3}{2 \, \sqrt{2} + e^{\left (-2 \, x\right )} + 3}\right ) - \frac{89 \, e^{\left (-2 \, x\right )} + 171 \, e^{\left (-4 \, x\right )} + 19 \, e^{\left (-6 \, x\right )} + 9}{16 \,{\left (12 \, e^{\left (-2 \, x\right )} + 38 \, e^{\left (-4 \, x\right )} + 12 \, e^{\left (-6 \, x\right )} + e^{\left (-8 \, x\right )} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+cosh(x)^2)^3,x, algorithm="maxima")

[Out]

-19/128*sqrt(2)*log(-(2*sqrt(2) - e^(-2*x) - 3)/(2*sqrt(2) + e^(-2*x) + 3)) - 1/16*(89*e^(-2*x) + 171*e^(-4*x)
+ 19*e^(-6*x) + 9)/(12*e^(-2*x) + 38*e^(-4*x) + 12*e^(-6*x) + e^(-8*x) + 1)

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Fricas [B]  time = 2.14874, size = 1935, normalized size = 37.94 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+cosh(x)^2)^3,x, algorithm="fricas")

[Out]

1/128*(152*cosh(x)^6 + 912*cosh(x)*sinh(x)^5 + 152*sinh(x)^6 + 456*(5*cosh(x)^2 + 3)*sinh(x)^4 + 1368*cosh(x)^
4 + 608*(5*cosh(x)^3 + 9*cosh(x))*sinh(x)^3 + 8*(285*cosh(x)^4 + 1026*cosh(x)^2 + 89)*sinh(x)^2 + 712*cosh(x)^
2 + 19*(sqrt(2)*cosh(x)^8 + 8*sqrt(2)*cosh(x)*sinh(x)^7 + sqrt(2)*sinh(x)^8 + 4*(7*sqrt(2)*cosh(x)^2 + 3*sqrt(
2))*sinh(x)^6 + 12*sqrt(2)*cosh(x)^6 + 8*(7*sqrt(2)*cosh(x)^3 + 9*sqrt(2)*cosh(x))*sinh(x)^5 + 2*(35*sqrt(2)*c
osh(x)^4 + 90*sqrt(2)*cosh(x)^2 + 19*sqrt(2))*sinh(x)^4 + 38*sqrt(2)*cosh(x)^4 + 8*(7*sqrt(2)*cosh(x)^5 + 30*s
qrt(2)*cosh(x)^3 + 19*sqrt(2)*cosh(x))*sinh(x)^3 + 4*(7*sqrt(2)*cosh(x)^6 + 45*sqrt(2)*cosh(x)^4 + 57*sqrt(2)*
cosh(x)^2 + 3*sqrt(2))*sinh(x)^2 + 12*sqrt(2)*cosh(x)^2 + 8*(sqrt(2)*cosh(x)^7 + 9*sqrt(2)*cosh(x)^5 + 19*sqrt
(2)*cosh(x)^3 + 3*sqrt(2)*cosh(x))*sinh(x) + sqrt(2))*log(-(3*(2*sqrt(2) - 3)*cosh(x)^2 - 4*(3*sqrt(2) - 4)*co
sh(x)*sinh(x) + 3*(2*sqrt(2) - 3)*sinh(x)^2 + 2*sqrt(2) - 3)/(cosh(x)^2 + sinh(x)^2 + 3)) + 16*(57*cosh(x)^5 +
342*cosh(x)^3 + 89*cosh(x))*sinh(x) + 72)/(cosh(x)^8 + 8*cosh(x)*sinh(x)^7 + sinh(x)^8 + 4*(7*cosh(x)^2 + 3)*
sinh(x)^6 + 12*cosh(x)^6 + 8*(7*cosh(x)^3 + 9*cosh(x))*sinh(x)^5 + 2*(35*cosh(x)^4 + 90*cosh(x)^2 + 19)*sinh(x
)^4 + 38*cosh(x)^4 + 8*(7*cosh(x)^5 + 30*cosh(x)^3 + 19*cosh(x))*sinh(x)^3 + 4*(7*cosh(x)^6 + 45*cosh(x)^4 + 5
7*cosh(x)^2 + 3)*sinh(x)^2 + 12*cosh(x)^2 + 8*(cosh(x)^7 + 9*cosh(x)^5 + 19*cosh(x)^3 + 3*cosh(x))*sinh(x) + 1
)

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Sympy [B]  time = 35.8442, size = 428, normalized size = 8.39 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+cosh(x)**2)**3,x)

[Out]

-19*sqrt(2)*log(4*tanh(x/2)**2 - 4*sqrt(2)*tanh(x/2) + 4)*tanh(x/2)**8/(128*tanh(x/2)**8 + 256*tanh(x/2)**4 +
128) - 38*sqrt(2)*log(4*tanh(x/2)**2 - 4*sqrt(2)*tanh(x/2) + 4)*tanh(x/2)**4/(128*tanh(x/2)**8 + 256*tanh(x/2)
**4 + 128) - 19*sqrt(2)*log(4*tanh(x/2)**2 - 4*sqrt(2)*tanh(x/2) + 4)/(128*tanh(x/2)**8 + 256*tanh(x/2)**4 + 1
28) + 19*sqrt(2)*log(4*tanh(x/2)**2 + 4*sqrt(2)*tanh(x/2) + 4)*tanh(x/2)**8/(128*tanh(x/2)**8 + 256*tanh(x/2)*
*4 + 128) + 38*sqrt(2)*log(4*tanh(x/2)**2 + 4*sqrt(2)*tanh(x/2) + 4)*tanh(x/2)**4/(128*tanh(x/2)**8 + 256*tanh
(x/2)**4 + 128) + 19*sqrt(2)*log(4*tanh(x/2)**2 + 4*sqrt(2)*tanh(x/2) + 4)/(128*tanh(x/2)**8 + 256*tanh(x/2)**
4 + 128) - 44*tanh(x/2)**7/(128*tanh(x/2)**8 + 256*tanh(x/2)**4 + 128) - 28*tanh(x/2)**5/(128*tanh(x/2)**8 + 2
56*tanh(x/2)**4 + 128) - 28*tanh(x/2)**3/(128*tanh(x/2)**8 + 256*tanh(x/2)**4 + 128) - 44*tanh(x/2)/(128*tanh(
x/2)**8 + 256*tanh(x/2)**4 + 128)

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Giac [A]  time = 1.29837, size = 96, normalized size = 1.88 \begin{align*} \frac{19}{128} \, \sqrt{2} \log \left (-\frac{2 \, \sqrt{2} - e^{\left (2 \, x\right )} - 3}{2 \, \sqrt{2} + e^{\left (2 \, x\right )} + 3}\right ) + \frac{19 \, e^{\left (6 \, x\right )} + 171 \, e^{\left (4 \, x\right )} + 89 \, e^{\left (2 \, x\right )} + 9}{16 \,{\left (e^{\left (4 \, x\right )} + 6 \, e^{\left (2 \, x\right )} + 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+cosh(x)^2)^3,x, algorithm="giac")

[Out]

19/128*sqrt(2)*log(-(2*sqrt(2) - e^(2*x) - 3)/(2*sqrt(2) + e^(2*x) + 3)) + 1/16*(19*e^(6*x) + 171*e^(4*x) + 89
*e^(2*x) + 9)/(e^(4*x) + 6*e^(2*x) + 1)^2